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NEET CHEMISTRYMedium

The correct order of dissociation energy of N2N_{2} and N2+N_{2}^{+} is:

A

N2>N2+N_{2} > N_{2}^{+}

B

N2=N2+N_{2} = N_{2}^{+}

C

N2+>N2N_{2}^{+} > N_{2}

D

None of the above

Step-by-Step Solution

Bond dissociation energy is the energy required to break one mole of bonds and is directly proportional to the bond order . According to Molecular Orbital Theory, the nitrogen molecule (N2N_{2}) has 14 electrons and a bond order of 3.0 . When N2N_{2} is ionized to form N2+N_{2}^{+}, one electron is removed from the highest occupied molecular orbital, which is a bonding orbital (σ2pz\sigma 2p_{z}) . This reduces the number of bonding electrons (NbN_{b}), resulting in a lower bond order of 2.5 for N2+N_{2}^{+} . Since N2N_{2} has a higher bond order than N2+N_{2}^{+}, it possesses a stronger bond and, consequently, a higher dissociation energy .

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